There is no general consensus regarding whether a ring should have a unity element or not. Many authors work with unital rings , and other does not essentially require unity. If we do not assume unity to be a necessary part of ring, lets call that structure , a "rng" (which may or may not have unity i.e. $1$), then many pathologies do occur.

I am hoping to get such pathologies (with at least one example) listed together on a page which I can keep as a record for future references.

I will post my list as an answer here, to get it started, and hope others will contribute to it, and make this post valuable.

  • 48,433
  • 4
  • 51
  • 124
Bhaskar Vashishth
  • 12,176
  • 1
  • 38
  • 79

4 Answers4


Note- In the following list, $R$ or $R_i$ will denote a "rng" and $S$ or $S_i $ will denote its "subrng"

  • Every Ideal is a subrng.
  • $S$ may not have unity even if $R$ has.

Example- $R=\Bbb{Z}$ and $S=2\Bbb{Z}$

  • $S$ has a unity where $R$ does not.

Example- $R= \Bbb{Z} \times \Bbb{2Z}$ and $S=\Bbb{Z}\times 0$

  • $S$ and $R$ may both have unity but they may not be same.

Example- $R$ = $\Bbb{Z_6}$ and $S= \{\bar{0}, \bar{2}, \bar{4}\}$ where $1_R=\bar{1}\ \text{and }\ 1_S=\bar{4}$.

  • If $R$ and $S$ have different unities, then $a\in S$ may be a unit in $S$ but not in $R$

Example- $R$ = $\Bbb{Z_6}$ and $S= \{\bar{0}, \bar{2}, \bar{4}\}$ then $a=\bar{2}$ is a unit in $S$ but not in $R$

  • $\text{Char}{(S)}< \text{Char}{(R)}$

Example- The above example works as $char(S)=3$ and $char(R)=6$

  • $a\in S $ may be a zero-divisor in $R$ but not in $S$.

Example- $R=\Bbb{Z}$ and $S=\{0\}$

  • Direct Product of "rngs" $R_1$ and $R_2$, i.e. $D=R_1 \times R_2$ is a ring iff both $R_1$ and $R_2$ are rings.
    In this case $1_D=(1_{R_1},1_{R_2}) $

  • Let $M=M_n(R)$, then there is no one-one correspondence between $2$- sided ideals of $R$ and $M$.

Example- Let $R=2\Bbb{Z}$, then $M=M_2{(2\Bbb{Z})}$. Then ideal $$J=\{\begin {pmatrix} a & b \\ c & d \end {pmatrix}\ |\ a,b,c,d \in 2\Bbb{Z} \text{with}\ a \in 4\Bbb{Z} \}$$.
Now it is easy to see that $J$ is an ideal strictly between $M_2{(4\Bbb{Z})}$ and $M_2(\Bbb{2Z})$ but there is no ideal between $\Bbb{4Z}$ and $2\Bbb{Z}$ in $\Bbb{Z}$.

  • If $a\in R$ is a zero-divisor in $R$ then $a+I$ need not be a zero-divisor in $R/I$.

Example- $\bar{3}$ is a zero divisor in $\Bbb{Z_6}$ but not in $\Bbb{Z_6}/I$ where $I=\{\bar{0}, \bar{2}, \bar{4}\}$

  • If $f$ is a "rng" homomorphism, then $f(1_R)$ need not equal $f(1_{R'})$

Example- Inclusion homomorphism $\{\bar{0}, \bar{2}, \bar{4}\} \hookrightarrow \Bbb{Z}_6$.

  • Jacobson radical for a "rng" is defined as, $$\text{rad}(R)= \{a \in R :\ Ra\ \text{is left quasi-regular}\}$$

  • Given an abelian group $(G,+)$ we can always make it into a "rng" by defining trivial multiplication i.e. $a.b=0$ for all $a,b\in G$, but that is not always true for non-finitely generated abelian groups, as if your group has the property that every element has finite order, but there is no upper bound on the orders of the elements, then it cannot be the additive abelian group of a ring with identity. The reason is that if there were such a ring structure with an identity $1$, then $1$ would have finite additive order $k$, and then for all $a$ in your group, $k\cdot a=(k\cdot1)a=0a=0$, which forces $a$ to have order at most $k$.

Example- Prüfer $p$-group $\mathbb Z(p^\infty)$ or the quotient group $\mathbb Q/\mathbb Z$.

Courtesy- Jonas Meyer's answer

Edit on 19/08/15-
For a commutative "rng", it is not necessary that every maximal ideal is prime, as $4\Bbb{Z}$ is maximal in $2\Bbb{Z}$ but not prime as $2.2\in 4\Bbb{Z}$ but $2\notin 4\Bbb{Z}$.

Bhaskar Vashishth
  • 12,176
  • 1
  • 38
  • 79
  • Your zero+divisor example is not correct, as $a\in S$ iff $a=0$, and either $0$ is a zero-divisor both places or neither. You should also add an example where $R$ has no unity but $S$ has one. – Tobias Kildetoft Jun 23 '15 at 10:58
  • $0$ is not a zero divisor in {$0$}, vacuously, as there is no nonzero element in $S$ – Bhaskar Vashishth Jun 23 '15 at 11:05
  • Ahh, good point (though still not a particularly good example I feel, as "being a zero-divisor" is usually not interesting unless one adds "non-zero"). – Tobias Kildetoft Jun 23 '15 at 11:07

I will add a little pathological thing to your list. Suppose $R$ is a commutative ring which may or may not have a unital element. Take $a\in R$. Then one may ask what is the ideal $(a)$?

A first answer is the following :

$$(a)=aR:=\{ar\mid r\in R\} $$

But if one does this, it may happen that $a\notin (a)$. For instance take the unital ring $A:=\mathbb{Z}[X]$ and define $R:=(X)=\{P\in A|P(0)=0\}$. Then $R$ is a ring without unity. Now take $a:=2X$ then :

$$(a)=\{2XP(X)|P\in R\} $$

I claim that $2X\notin (2X)$ (the valuation of any non-zero element of $(2X)$ is $>1$).

Hence one sees that $aR$ is not the smallest ideal containing $a$.

A second answer is the following :

$$(a)=\langle a,ar\mid r\in R\rangle $$

That is the additive group generated by $a$ and all elements $ar$ where $r\in R$. I claim that this is the smallest ideal containing $a$ but in this case $(a)\neq aR$ which is (IMHO) very disturbing...

Clément Guérin
  • 11,049
  • 1
  • 14
  • 38
  • nice. So basically which answer is correct to use? It is no rule that $a$ must be in ($a$), so it is safe to assume first definition. Same with second, so basically it depends on what we want out of $(a)$ – Bhaskar Vashishth Jun 23 '15 at 11:46
  • 1
    @BhaskarVashishth I actually had a little (amical) discussion about this in MSE. I would have though that the first definition is better. However, you really want $(a)$ to be the smallest ideal containing $a$ (with more than one element, it is basically the only good definition possible). So the second definition is better. For what it's worth I think I remember that Wikipedia says that the second is better... – Clément Guérin Jun 23 '15 at 11:51
  • 1
    @BhaskarVashishth, by the way, Batominovsky's answer also says that the second definition is better. His answer (a little technical) provides a good justification of this. – Clément Guérin Jun 23 '15 at 11:54

I think, one of the most useful concepts about rings is something I would like to call the "unital closures." (I refuse to use the term "rngs" and, in my post, the term "rings" includes both unital rings and nonunital rings.) For any ring $R$, the "unital closure" is the ring $\hat{R}:=\mathbb{Z}\oplus R$ with the addition $(k,r)+(l,s):=(k+l,r+s)$ and the multiplication $(k,r)\cdot(l,s):=(kl,ks+lr+rs)$ for $k,l\in\mathbb{Z}$ and $r,s\in R$. Clearly, $\hat{R}$ is a unital ring with the multiplicative identity $\left(1,0_R\right)$.

For subsets $A,B$ of $R$, write $A\cdot B$ for the set containing all sums of the form $\sum_{i=1}^n \,a_ib_i$, where $n\in\mathbb{N}$, $a_i\in A$, and $b_i\in B$ for all $i$. A left ideal of $R$ is then an additive abelian subgroup $L$ of $R$ such that $R\cdot L\subseteq I$. If $L$ is generated by a subset $X$ of $R$, then $L$ consists of all elements of the form $\sum_{i=1}^p\,k_ix_i+\sum_{j=1}^q\,r_jy_j$, where $p,q\in\mathbb{N}$, $k_i\in\mathbb{Z}$, $r_j\in R$, and $x_i,y_j\in X$ for every $i,j$. Hence, unlike in the case of unital rings, it is possible that $L\neq R\cdot X$. A two-sided ideal of $R$ is a left ideal $I$ of $R$ such that $I\cdot R \subseteq I$. If $I$ is generated by a subset $X$ of $R$, then $I$ consists of all elements of the form $\sum_{i=1}^p\,k_ix_i+\sum_{j=1}^q\,r_jy_j+\sum_{k=1}^m\,z_ks_k+\sum_{l=1}^n\,t_lw_lu_l$, where $p,q,m,n\in\mathbb{N}$, $k_i\in\mathbb{Z}$, $r_j,s_k,t_l,u_l\in R$, and $x_i,y_j,z_k,w_l\in X$ for every $i,j,k,l$. Again, it is possible that $I\neq R\cdot X\cdot R$, unless $R$ is unital.

A ring homomorphism $\phi:R\to S$ for rings $R$ and $S$ is defined to only be additive and multiplicative. If $R$ and $S$ are unital, then $\phi$ is "unitary" if $\phi\left(1_R\right)=1_S$. Isomorphism theorems for unitary ring homomorphisms carry over to isomorphisms theorems of homomorphisms of potentially nonunital rings. Furthermore, $R$ can be viewed as a subring of $\hat{R}$ via the embedding $r\mapsto (0,r)$ for all $r\in R$. Anyway, any ring homomorphism $\phi:R\to S$ can be extended uniquely to a unitary ring homomorphism $\hat{\phi}:\hat{R}\to\hat{S}$ via $\hat{\phi}\big((k,r)\big):=\big(k,\phi(r )\big)$ for every $k\in\mathbb{Z}$ and $r\in R$. Hence, in a sense, $\mathrm{Hom}(R,S)=\mathrm{Hom}_{\text{unitary}}\left(\hat{R},\hat{S}\right)$. A good thing about taking nonunital rings into account is that the category $\mathbf{Rings}$ of all rings together with ring homomorphisms has a zero object (the zero ring), but the category $\mathbf{URings}$ of all unital rings together with unitary ring homomorphisms has no zero objects, but an initial object ($\mathbb{Z}$) and a terminal object (again, the zero ring).

A left $R$-module $M$ is defined in the usual way, and if $R$ is unital, we say that $M$ is "unitary" if $1_R\cdot m=m$ for all $m\in M$. Homomorphisms of left $R$-modules are also as usual. I shall denote by $\mathbf{LMod}(R )$ the category of left $R$-modules together with homomorphisms of left $R$-modules, and by $\mathbf{LUMod}(U )$ the category of unitary left $U$-modules together with homomorphisms of left $U$-modules, given that $U$ is a unital ring. Obviously, $\mathbf{LUMod}(U )$ is a full subcategory of $\mathbf{LMod}(U)$ for every unital ring $U$.

Another essential thing about the unital closure $\hat{R}$ of a ring $R$ is that $\mathbf{LMod}(R )$ is identical to $\mathbf{LUMod}\left(\hat{R}\right)$ (since every left $R$-module $M$ is a left $\hat{R}$-module via $(k,r)\cdot m :=km+r\cdot m$ for $k\in\mathbb{Z}$, $r\in R$, and $m\in M$, and every left $\hat{R}$-module $N$ is a left $R$-module via $r\cdot n:=(0,r)\cdot n$ for all $r\in R$ and $n\in N$). Hence, many properties of unitary modules over a unital ring can be generalized to modules over a ring. For example, you know that a unitary free module over a unitary ring $U$ is a direct sum of copies of $U$. What about a free module over $R$? Well, as $\hat{R}$ is unital, a free module over $R$ then is a unitary free module over $\hat{R}$, hence it will be a direct sum of copies of $\hat{R}=\mathbb{Z}\oplus R$. Another example is that $\mathbf{LMod}(R )$ has enough projectives and enough injectives, which is due to the fact that $\mathbf{LUMod}(U)$ has enough projectives and enough injectives.

Now, I return again to Clément Guérin's answer about ideals of rings. For a ring $R$, a left ideal of $R$ is actually a left $\hat{R}$-submodule of $R$, where $R$ is given the left $\hat{R}$-module structure $(k,r)\cdot s:=ks+rs$ for $k\in\mathbb{Z}$ and $r,s\in R$. So, the left ideal $\langle a \rangle$ generated by $a\in R$ will be $\left\{ka+ra\,|\,k\in\mathbb{Z}\text{ and }r\in R\right\}$, which is equal to $R\cdot a$ if $R$ is unital.

PS: I omitted the discussion of right ideals and right modules for a trivial reason.

I just remembered a problem I dealt with recently.

Let $R$ be a ring and $n\in\mathbb{N}$. What are all left ideals and two-sided ideals of the ring of matrices $S:=\mathrm{Mat}_{n\times n}(R)$?

Assume that $R$ is unital. Then, there is a one-to-one correspondence between the set $\mathcal{L}\left(R^n\right)$ of all left $R$-submodules of the unitary $R$-module $R^n$ and the set $\mathcal{L}(S)$ of all left ideals of $S$ which associates each $V\in\mathcal{L}\left(R^n\right)$ with the left ideal of $S$ consisting of matrices $\left[v_{i,j}\right]_{i,j\in[n]}$, where $[n]:=\{1,2,\ldots,n\}$, and $\left(v_{i,1},v_{i,2},\ldots,v_{i,n}\right) \in V$ for every $i\in[n]$. There is also a one-to-one correspondence between the set $\mathcal{T}(R)$ of two-sided ideal of $R$ and the set $\mathcal{T}(S)$ of two-sided ideals of $S$. This correspondence associates $I\in\mathcal{T}(R)$ with $\text{Mat}_{n\times n}(I)$.

Now, what happens if $R$ is nonunital? I don't know the answer, and I suspect that it is an open question. I don't even know the answer when $R$ is a trivial ring (i.e., an additive abelian group $R$ with the trivial multiplication: $r\cdot s:=0_R$ for all $r,s\in R$).

  • 48,433
  • 4
  • 51
  • 124

It seems like no one has mentioned the pathology that first comes to my mind for rings without identity:

A ring without identity may not have maximal (left/right/two-sided) ideals.

The fact that a ring with identity must have maximal ideals is a sort of very weak finiteness condition. Of course, it is not a spectacular finiteness condition since there are rings without identity that have maximal ideals anyway (even finite rings without identity) but still it is a special condition.

Example- $(\Bbb{Q},+,\circ)$ where $\circ$ is trivial multiplication has no maximal ideals as ideals in this rng are same as subgroups of $(\Bbb{Q},+)$ and we can prove that $(\Bbb{Q},+)$ has no maximal subgroups.

Bhaskar Vashishth
  • 12,176
  • 1
  • 38
  • 79
  • 141,564
  • 15
  • 151
  • 364
  • ohh yes how can I forget that. – Bhaskar Vashishth Jun 23 '15 at 12:26
  • I have edited it with example. Please see if it fine with you or rollback. Thanks – Bhaskar Vashishth Jun 23 '15 at 12:29
  • Don't the rationals have an identity element though (namely 1)? – Zach Effman Jun 23 '15 at 21:22
  • 4
    @ZachEffman by "trivial multiplication" I believe they mean the operation $a \circ b = 0$ for all $a, b \in \mathbb{Q}$, not the usual multiplication in $\mathbb{Q}$. You can take any abelian group $(G, +, 0)$ and give it this multiplication operation, and you'll get a (commutative) ring...but not a unital ring unless $G$ has only one element. – mathmandan Jun 23 '15 at 22:43